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Home/ Questions/Q 8727093
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T08:22:33+00:00 2026-06-13T08:22:33+00:00

I have a quick question in the program I’m trying to create. I can’t

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I have a quick question in the program I’m trying to create. I can’t seem to figure out how I can carry through the value that someone inputs for my variable of “denom” so that it can be successfully used in the function I’ve created. Help would be much appreciated.

#include <stdio.h>

//Global Variables
int num, denom;
void simplify(int *numerator, int *denominator);

int main () {
    int num1, denom1;
    //Prompt User as to what program is
    printf("Fraction Simplifier\n");
    printf("===================\n");

    //Ask User for Numerator and Denominator
    printf("Numerator: ");
    scanf("%d", &num);
    printf("Denominator: ");
    scanf("%d", &denom);

    //Call Function
    simplify(&num1, &denom1);

    //Display final output
    printf("%d / %d = %d / %d\n", num, denom, num1, denom1);

    return 0;
}

//Simplify function
void simplify(int *numerator, int *denominator)
{

    int num1, denom1;
    num1 = 1;
    num = num1;
    denom = denom1;

    num1 = num1 / num1;
    denom1 = denom1 / num1;
    *numerator = num1 ;
    *denominator = denom1;

}
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  1. Editorial Team
    Editorial Team
    2026-06-13T08:22:34+00:00Added an answer on June 13, 2026 at 8:22 am

    Your simplify() method doesn’t even try to use the passed values.

    It boils down to *numerator = 1 and *denominator = is undefined (since denom1 is never initialized). i.e. it sets “out values” to “constants”. You probably meant to do something like:

    int num1 = *numerator;
    int denom1 = *denominator;
    
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